B. The Meeting Place Cannot Be Changed The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost build…

Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at hom…

题意:给定x轴上有n个点,每一个点都有一个权值,让在x轴上选一个点,求出各点到这个点的距离的三次方乘以权值最小. 析:首先一开始我根本不会三分,也并没有看出来这是一个三分的题目的,学长说这是一个三分的题,我就百度了一下什么是三分算法,一看感觉和二分差不多,当然就是和二分差不多,也是慢慢缩短范围. 这个题也这样,在最左端和最右端不断的三分,直到逼进那个点,刚开始我设置的误差eps是10负8,但是TLE了,我以为是太小,三分数太多,然后我又改成10负6还是TLE,我又失望了,干脆我不用误差了,我让它…

Party All the Time Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4282    Accepted Submission(s): 1355 Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits wil…

题意:n个人,都要去參加活动,每一个人都有所在位置xi和Wi,每一个人没走S km,就会产生S^3*Wi的"不舒适度",求在何位置举办活动才干使全部人的"不舒适度"之和最小,并求最小值. 思路:首先能够得出最后距离之和的表达式最多仅仅有两个极点, 更进一步仅仅有一个极点,否则无最小值. 那么我们就可用三分法或者二分法求解.即对原函数三分或对导数二分就可以. #include<cstdio> #include<cstring> #inclu…

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6979 Accepted Submission(s): 2181 Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717 说明下为啥满足三分: 设y=f(x) (x>0)表示任意两个点的距离随时间x的增长,距离y的变化.则f(x)函数单调性有两种:1.先单减,后单增.2.一直单增. 设y=m(x) (x>0)表示随时间x的增长,所有点的最大距离y的变化.即m(x)是所有点对构成的f(x)图像取最上面的部分.则m(x)的单调性也只有两种可能:1.先单减,后单增.2.一直单增. 这个地方的证明可以这样:假如时刻t1…

NPY and shot Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description NPY is going to have a PE test.One of the test subjects is throwing the shot.The height of NPY is H meters.He can throw the shot at t…

http://acm.hdu.edu.cn/showproblem.php?pid=4717 大致题意:给出每一个点的坐标以及每一个点移动的速度和方向. 问在那一时刻点集中最远的距离在全部时刻的最远距离中最小. 比赛时一直以为是计算几何,和线段相交什么的有关.赛后队友说这是道三分,细致想了想确实是三分.试着画绘图发现它是一个凸性函数,存在一个最短距离. 然后三分时间就能够了. #include <stdio.h> #include <iostream> #include <m…

我比较快速的想到了三分,但是我是从0到2*pi区间进行三分,并且漏了一种点到边距离的情况,一直WA了好几次 后来画了下图才发现,0到2*pi区间内是有两个极值的,每个半圆存在一个极值 以下是代码 #include <cstdio> #include <cmath> #include <algorithm> #define pi acos(-1.0) using namespace std; typedef struct { double x; double y; }po…